Periodic table
Problem
Given the following reaction:
\qquad \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH
4

+2O
2

→CO
2

+2H
2

Ostart text, C, H, end text, start subscript, 4, end subscript, plus, 2, start text, O, end text, start subscript, 2, end subscript, right arrow, start text, C, O, end text, start subscript, 2, end subscript, plus, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text
How many grams of \text{CO}_2CO
2

start text, C, O, end text, start subscript, 2, end subscript will be produced from 12.0 \text{ g}12.0 g12, point, 0, start text, space, g, end text of \text{CH}_4CH
4

start text, C, H, end text, start subscript, 4, end subscript and 133 \text{ g}133 g133, start text, space, g, end text of \text{O}_2O
2

start text, O, end text, start subscript, 2, end subscript?
grams (round to three significant figures)

33g of CO2

Explanation:

Step 1:

The balanced equation for the reaction.

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of CO2 produced from the balanced equation.

This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of CO2 = 12 + (2x16) = 44g/mol.

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 44g of CO2.

Step 3:

Determination of limiting reactant. This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 12g of CH4 will react with = (12 x 64)/16 = 48g of O2.

From the above illustration, a lesser mass of O2 i.e 48g than what was given i.e 133g is required to react completely with 12g of CH4.

Therefore, CH4 is the limiting reactant and O2 is the excess reactant.

Step 4:

Determination of the mass of CO2 produced from the reaction.

Here, we shall use the limiting reactant, because it will give the maximum yield of CO2 as all of it is used up in reaction.

This is illustrated as follow:

From the balanced equation above,

16g of CH4 reacted to produce 44g of CO2.

Therefore, 12g of CH4 will react to produce = (12 x 44)/16 = 33g of CO2.

From the calculations made above, 33g of CO2 will be produced from the reaction of 12g of CH4 and 133g of O2.

answer; i believe the correct answer is (solid gold);

l dont know l am s sorry